Answer: 4.5 x 10e-7
Explanation: 450 x 1e+9 = correct answer
Multiply amount of nanometers by 1e+9 to get the approximate result in meters.
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
There is no reaction.
<em>Molecular equation
:</em>
K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)
<em>Ionic equation
:</em>
2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)
<em>Net ionic equation
:</em>
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>
<em>All ions cancel</em>. There is no net ionic equation.
Answer:
One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles
